Poinsot construction

Fig. 1: Poinsot's construction

The Poinsot'sche construction according Louis Poinsot models the movement of the force-free top as a slip-free rolling of the Energieellipsoids on a fixed invariable level , see FIG. 1.

The angular velocity plotted in the center of mass ends in the pole ( Greek πόλος pólos "axis"). This moves in the fixed body system on closed curves, the polhodia ("pole paths" from ὁδός hodós "way, path, road"), which lie on the energy ellipsoid or poinsotellipsoid . Depending on whether the polhodies enclose the main axis of inertia with the smallest or the largest main moment of inertia, the polhodies are called epi or pericycloid . The polhody in Fig. 1 is epicycloidal. In the inertial system fixed in space, the angular velocity in the pole touches the invariable plane and traces the Herpolhodien ("meandering paths of the pole" from ἕρπω hérpo "creeping"). The invariable plane touches the poinsotellipsoid at all times.

The elements mentioned form the Poinsot construction and their time course defines the Poinsot movement . With Poinsot's construction, the investigation of the rotational movement of rigid bodies becomes a geometrical task.

Animations
Epicycloidal Movement	Pericycloidal movement	Movement close to the separatrix, see Dschanibekow effect
Unlike in the animations, the angular momentum relates to the center of mass.

description

Poinsot's construction considers a force-free top that rests in its center of mass. In addition to weightlessness, a force-free body can be realized in a gravitational field by being rotatable in its center of gravity, for example by gimbaling it.

The expansion of the energy ellipsoid is constant because it is determined by the rotational energy , which in the case of a force-free top is an integral of its movement, since no work is done due to the lack of external forces. Where the current angular velocity touches the energy ellipsoid, the current angular momentum is perpendicular to the tangential plane. The angular momentum is unchangeable in the force-free top and thus the tangential planes are parallel to each other or coincide during the movement.

The component of the angular velocity in the direction of the angular momentum always remains the same. Because with the force-free rotational movement of a body, both its rotational energy E _rot and its angular momentum are preserved. The former is calculated from the latter by scalar multiplication with the angular velocity: ${\displaystyle {\vec {L}}}$

${\displaystyle 2E_{\text{rot}}={\vec {L}}\cdot {\vec {\omega }}=L\omega \cos \varphi \quad \rightarrow \quad \omega \cos \varphi ={\frac {2E_{\text{rot}}}{L}}}$

Therein is L , the amount of angular momentum, ω the amount of the rotational speed, and φ included by angular momentum and speed of rotation angle. On the right side of the last equation there is a constant of the rotary motion, which is why the left side, the portion of the angular velocity in the direction of the angular momentum, is also constant. Said part determines the distance of the tangential plane from the center of mass. This lies at the origin O and its base point on the tangential plane is A. Then this fixed component of the angular velocity is the distance OA. Thus the tangential plane to the Poinsotellipsoid in the pole is fixed and is called the invariable plane (green plane in Fig. 1 ).

In spite of the fact that the angular velocity OA is constant, the pole beam AP does not rotate at a constant rotational speed around the axis OA, because the pole not only migrates in the plane, but also on the poinsotellipsoid.

A particle of the rigid body located in the pole is momentarily still because it lies on the current axis of rotation, which goes through the center of mass at rest.

If the body does not orbit around one of its main axes of inertia, then at most one of the angular velocities ω _1,2,3 can be zero. The Euler equations gyro show that then never all three can also disappear from the angular accelerations. Accordingly, the pole does not stop on the polhodia and herpolhodia, or even reverses its direction of movement.

Polhodies

Epi- and pericycloid polhodies

Fig. 2: Polhodia on the poinsotellipsoid (gray) and twist ellipsoid (yellow).

On the one hand, because of the conservation of energy, the angular velocity lies on the poinsotellipsoid (gray in Fig. 2). On the other hand, because of the conservation of angular momentum, it also touches the helix ellipsoid , which in the body-fixed system contains the end points of all angular velocity vectors that lead to the same angular momentum square (yellow). The polhodies are the intersection curves of these two ellipsoids and, as such, are circular, elliptical or taco- shaped, closed curves which, like the ellipsoids , are symmetrical about all three main planes generated by the main axes of inertia. The polhodies shown in red in Fig. 2 are called epicycloidal polhodes after Arnold Sommerfeld and Felix Klein . For them, L ² <2Θ ₂ E _rot , where E _{rot is} the rotational energy, L is the amount of angular momentum and Θ _{2 is} the mean main moment of inertia. The curves drawn in blue are the pericycloid polhodies, where 2Θ ₂ E _red  <  L ². Between the epi- and pericycloidic polhodies lies the separating polhody or #Separatrix (black), which arises in _red at L ² = 2Θ ₂ E and can be thought of as being composed of two ellipses.

Points of contact of the ellipsoids

For a given rotational energy, the smallest possible twist ellipsoid touches the poinsotellipsoid at the end points of the major axis. This situation corresponds to a uniform rotation around the main axis of inertia with the smallest main moment of inertia, because the lengths of the axes are inversely proportional to the main moments of inertia. Here the angular momentum has the minimum amount that is compatible with the rotational energy. When the largest possible twist ellipsoid touches the poinsotellipsoid at the end points of the smallest axis, a uniform rotation takes place around the main axis of inertia with the largest main moment of inertia, and the angular momentum has reached the maximum amount that is compatible with the rotational energy.

Mathematically it is expressed as follows: If the body rotates with the angular velocity ω _k around the kth main axis with the main moment of inertia Θ _k , then it has the angular momentum L _k  = Θ _k ω _k and the rotational energy

${\displaystyle E_{\text{rot}}={\frac {1}{2}}\Theta _{k}\omega _{k}^{2}={\frac {L_{k}^{2}}{2\Theta _{k}}}\quad \rightarrow \quad L_{k}={\sqrt {2E_{\text{rot}}\Theta _{k}}}}$

The amount of angular momentum is greatest or smallest when the body rotates around its main axis with the largest or smallest main moment of inertia.

Rotating and oscillating movements

A rotation about the 1-axis takes place on the epicycloidal polhodia and the angle of rotation about this axis is unlimited. In the pericycloidal polhodia the angle of rotation about the 1-axis fluctuates between two extreme values. Correspondingly, the epicycloid movements are called rotating and the pericycloid movements are called oscillating .

Stability considerations

If the pole is only in the vicinity but not on the largest or smallest axis, it also remains in their vicinity, because the polhodia enclose these endpoints. This is different on the separatrix, where a pole that is close but not on the central axis is significantly removed from its initial position on an epi- or pericycloidal pole and does not wrap around the axis either. The largest and smallest axes thus mark stable axes of rotation, whereas the middle axis of rotation is an unstable one.

In the case of strongly flattened or very slim ellipsoids, even a small impact can move the pole far away from the main axis of inertia, even if the movement takes place around one of the stable axes. Thus, even a stable axis of rotation can appear unstable if the main moments of inertia are very different. A measure of the stability of the axes of rotation can be derived from the axial ratios of the ellipses, as which the polhodies appear when viewed from the direction of the main axes of inertia. The angular velocities satisfy the two equations

${\displaystyle {\begin{aligned}2E_{\text{rot}}=&\Theta _{1}\omega _{1}^{2}+\Theta _{2}\omega _{2}^{2}+\Theta _{3}\omega _{3}^{2}\\L^{2}=&\Theta _{1}^{2}\omega _{1}^{2}+\Theta _{2}^{2}\omega _{2}^{2}+\Theta _{3}^{2}\omega _{3}^{2}\,.\end{aligned}}}$

Projection of the intersection curves in the direction of one of the main axes of inertia onto a plane perpendicular to it takes place by eliminating the angular velocity component in the direction of the axis, which is based on the equations

${\displaystyle {\begin{aligned}L^{2}-2\Theta _{1}E_{\text{rot}}=&\Theta _{2}(\Theta _{2}-\Theta _{1})\omega _{2}^{2}+\Theta _{3}(\Theta _{3}-\Theta _{1})\omega _{3}^{2}\\2\Theta _{2}E_{\text{rot}}-L^{2}=&\Theta _{1}(\Theta _{2}-\Theta _{1})\omega _{1}^{2}+\Theta _{3}(\Theta _{2}-\Theta _{3})\omega _{3}^{2}\\2\Theta _{3}E_{\text{rot}}-L^{2}=&\Theta _{1}(\Theta _{3}-\Theta _{1})\omega _{1}^{2}+\Theta _{2}(\Theta _{3}-\Theta _{2})\omega _{2}^{2}\end{aligned}}}$

leads. The first and third equations only have positive coefficients, which is why they describe ellipses that represent the axial relationships

${\displaystyle s_{1}={\sqrt {\frac {\Theta _{2}(\Theta _{2}-\Theta _{1})}{\Theta _{3}(\Theta _{3}-\Theta _{1})}}},\qquad s_{3}={\sqrt {\frac {\Theta _{2}(\Theta _{3}-\Theta _{2})}{\Theta _{1}(\Theta _{3}-\Theta _{1})}}}}$

exhibit. The stability decreases the further the ratios move away from one, and becomes greatest when the top is symmetrical with respect to the 1- or 3-axis, because then s ₁ = 1 or s ₃ = 1.

Separatrix

Fig. 3: Path of a point on the 2-axis (red) around the angular momentum axis (vertical line) along a loxodrome

On the separatrix, L ² = 2Θ ₂ E is _red and the second of the above ellipse equations is defined according to

${\displaystyle {\begin{aligned}2\Theta _{2}E_{\text{rot}}-L^{2}=0=&\Theta _{1}(\Theta _{2}-\Theta _{1})\omega _{1}^{2}+\Theta _{3}(\Theta _{2}-\Theta _{3})\omega _{3}^{2}\\\rightarrow \omega _{1}=&\pm {\sqrt {\frac {\Theta _{3}(\Theta _{3}-\Theta _{2})}{\Theta _{1}(\Theta _{2}-\Theta _{1})}}}\omega _{3}\end{aligned}}}$

two straight lines through the origin in the 1-3 plane. The planes spanned by these straight lines and the 2-axis contain the separatrix, which consists of ellipses as planar sections of an ellipsoid (black in Fig. 2 ). The movement shows that a point on the 2-axis on a loxodrome rotates infinitely often around the angular momentum axis at a constant speed, see Fig. 3 and movement on the separatrix . The pole asymptotically approaches the intersection of the two ellipses on the 2-axis, but never reaches it.

Herpolhodes

Fig. 4: Herpolhodia with epi- and pericycloid movement as well as movement on the separatrix. Dashed: polhody projected into the invariable plane at a point in time.

The Herpolhodien trace the path of the pole in the invariable plane. Because the part of the angular velocity that is perpendicular to the angular momentum, the pole beam AP, like the angular velocity itself, fluctuates between two extreme values, the Herpolhodes lie between two concentric circles around the base point A of the center of mass on the invariable plane, see Fig. 4. The Herpolhodes are mostly not closed, as in Fig. 4, after which the top no longer needs to return to its initial position. Despite being named as a winding path, the Herpolhodien have no turning points and no peaks. The center of curvature is always on the side of the foot point A .

proof

The angular velocity is expressed in the main axis system fixed to the body and is used to calculate the rates of the basic vectors according to ${\displaystyle {\hat {g}}_{1,2,3}}$ ${\displaystyle {\vec {\omega }}=\sum _{i=1}^{3}\omega _{i}{\hat {g}}_{i}\quad {\text{und}}\quad {\dot {\hat {g}}}_{i}={\vec {\omega }}\times {\hat {g}}_{i}.}$ Rotations away from the main axes are considered, so that at most one of the angular velocities ω 1,2,3 should be zero. The speed of the pole is and is: ${\displaystyle {\dot {\vec {\omega }}}}$ ${\displaystyle {\dot {\vec {\omega }}}=\sum _{i=1}^{3}({\dot {\omega }}_{i}{\hat {g}}_{i}+\omega _{i}{\dot {\hat {g}}}_{i})=\sum _{i=1}^{3}({\dot {\omega }}_{i}{\hat {g}}_{i}+{\vec {\omega }}\times \omega _{i}{\hat {g}}_{i})=\sum _{i=1}^{3}{\dot {\omega }}_{i}{\hat {g}}_{i}.}$ Because of the angular accelerations result from Euler's gyroscopic equations : ${\displaystyle {\vec {\omega }}\times {\vec {\omega }}={\vec {0}}.}$ ${\displaystyle {\begin{aligned}{\dot {\omega }}_{1}=&-{\frac {\Theta _{3}-\Theta _{2}}{\Theta _{1}}}\omega _{2}\omega _{3}=-p_{1}\omega _{2}\omega _{3}\quad {\text{mit}}\quad p_{1}:={\frac {\Theta _{3}-\Theta _{2}}{\Theta _{1}}}>0\\{\dot {\omega }}_{2}=&{\frac {\Theta _{3}-\Theta _{1}}{\Theta _{2}}}\omega _{3}\omega _{1}=p_{2}\omega _{1}\omega _{3}\quad {\text{mit}}\quad p_{2}:={\frac {\Theta _{3}-\Theta _{1}}{\Theta _{2}}}>0\\{\dot {\omega }}_{3}=&-{\frac {\Theta _{2}-\Theta _{1}}{\Theta _{3}}}\omega _{1}\omega _{2}=-p_{3}\omega _{1}\omega _{2}\quad {\text{mit}}\quad p_{3}:={\frac {\Theta _{2}-\Theta _{1}}{\Theta _{3}}}>0\end{aligned}}}$ Because, according to the assumption, at most one of the angular velocities is zero, all three angular accelerations can never vanish at the same time, so that the pole can never stop and the Herpolhodia therefore have no peaks. The ratios p 1,2,3 are all in the open interval (0,1), because the principal moments of inertia satisfy the triangle inequalities, and p 2 is the largest, because: ${\displaystyle {\begin{aligned}p_{2}-p_{1}=&{\frac {\Theta _{3}-\Theta _{1}}{\Theta _{2}}}-{\frac {\Theta _{3}-\Theta _{2}}{\Theta _{1}}}={\frac {(\Theta _{2}-\Theta _{1})(\Theta _{1}+\Theta _{2}-\Theta _{3})}{\Theta _{1}\Theta _{2}}}>0\\p_{2}-p_{3}=&{\frac {\Theta _{3}-\Theta _{1}}{\Theta _{2}}}-{\frac {\Theta _{2}-\Theta _{1}}{\Theta _{3}}}={\frac {(\Theta _{3}-\Theta _{2})(\Theta _{2}+\Theta _{3}-\Theta _{1})}{\Theta _{2}\Theta _{3}}}>0.\end{aligned}}}$ The acceleration of the pole is ${\displaystyle {\begin{aligned}{\ddot {\vec {\omega }}}=&\sum _{i=1}^{3}({\ddot {\omega }}_{i}{\hat {g}}_{i}+{\dot {\omega }}_{i}{\dot {\hat {g}}}_{i})=\sum _{i=1}^{3}({\ddot {\omega }}_{i}{\hat {g}}_{i}+{\vec {\omega }}\times {\dot {\omega }}_{i}{\hat {g}}_{i})=\sum _{i=1}^{3}{\ddot {\omega }}_{i}{\hat {g}}_{i}+{\vec {\omega }}\times {\dot {\vec {\omega }}}\end{aligned}}}$ with the angular backs ${\displaystyle {\begin{aligned}{\ddot {\omega }}_{1}=&-p_{1}({\dot {\omega }}_{2}\omega _{3}+\omega _{2}{\dot {\omega }}_{3})=p_{1}\omega _{1}(p_{3}\omega _{2}^{2}-p_{2}\omega _{3}^{2})\\{\ddot {\omega }}_{2}=&p_{2}({\dot {\omega }}_{3}\omega _{1}+\omega _{3}{\dot {\omega }}_{1})=-p_{2}\omega _{2}(p_{3}\omega _{1}^{2}+p_{1}\omega _{3}^{2})\\{\ddot {\omega }}_{3}=&-p_{3}({\dot {\omega }}_{1}\omega _{2}+\omega _{1}{\dot {\omega }}_{2})=p_{3}\omega _{3}(p_{1}\omega _{2}^{2}-p_{2}\omega _{1}^{2}).\end{aligned}}}$ After elementary transformations it results ${\displaystyle {\ddot {\vec {\omega }}}={\begin{pmatrix}-\omega _{1}[p_{3}(1-p_{1})\omega _{2}^{2}+p_{2}(p_{1}+1)\omega _{3}^{2}]\\\omega _{2}[(p_{3}(1-p_{2})\omega _{1}^{2}-p_{1}(p_{2}+1)\omega _{3}^{2}]\\\omega _{3}[(p_{1}(p_{3}+1)\omega _{2}^{2}+p_{2}(1-p_{3})\omega _{1}^{2}]\end{pmatrix}}}$ The square brackets in the first and third components are positive and because only at most one of the angular velocities should be zero, the pole acceleration never disappears. The cross product with the pole velocity gives: ${\displaystyle {\begin{aligned}{\dot {\vec {\omega }}}\times {\ddot {\vec {\omega }}}={\begin{pmatrix}\omega _{2}\omega _{3}[(p_{2}-p_{3})\omega _{1}^{2}+p_{1}(p_{2}+1)\omega _{3}^{2}+p_{1}(p_{3}+1)\omega _{2}^{2}]\\-\omega _{1}\omega _{3}[(p_{1}+p_{3})\omega _{2}^{2}+p_{2}(1-p_{3})\omega _{1}^{2}+p_{2}(p_{1}+1)\omega _{3}^{2}]\\\omega _{1}\omega _{2}[p_{3}(1-p_{1})\omega _{2}^{2}+p_{3}(1-p_{2})\omega _{1}^{2}+(p_{2}-p_{1})\omega _{3}^{2}]\end{pmatrix}}\end{aligned}}}$ The cross product disappears when the pole acceleration and velocity are parallel and thus possibly a turning point in the Herpolhodie occurs. However, the square brackets are all positive, so that not all three components can disappear at once. The Herpolhodia cannot have a turning point.

Symmetrical tops

With symmetrical gyroscopes, two main moments of inertia coincide, so that the poinsotellipsoid and the twist ellipsoid are rotationally symmetrical . The polhodies and the herpolhodes then become circles. All angular velocities on the polhodia and on the herpolhodia form a cone, the track cone and polar cone , which represent circular cones in the symmetrical top . The symmetrical, elongated, prolate top can only move epicycloidically, the symmetrical, flattened, oblate top only pericycloidically. If the polhodial circle of the prolate top is folded into the invariable plane, it lies outside the herpolhodial circle. Any point on the folded circle of polhodia drives an epicycloid when rolling on the circle of herpolhodes . The circle of polhodes of the oblate top, folded into the invariable plane, on the other hand, rolls on the inside of the circle of herpolhodes, which the circle of polhodes encloses, and a point on it draws a pericycloid . This motivates the designation of the movement as epi- or pericycloid. The case can never arise in which the rolling circle of polhodes lies within the fixed circle of herpolhodes and the movement should be called hypocycloid accordingly .

Lagrange top

Poinsot's construction can also be transferred to Lagrange tops . The Lagrange top is a symmetrical, heavy top with the center of mass lying on the figure axis and with a support point. In heavy gyroscopes with a support point, the angular momentum moves in a plane that is perpendicular to the weight force, and the distance between this plane and the origin is constant, as it is an integral of the movement. Since the total energy of the top is constant, but not its rotational energy, the Poinsot ellipsoid has an expansion that increases and decreases in the opposite direction to the positional energy . The polhodies lie in a plane that is perpendicular to the figure axis and has the distance ω ₃ from the support point, because the axial angular velocity ω ₃ is constant in the Lagrange top.

General case

In the Lagrangian top, the Herpolhodia are generally spherical curves that run on the surface of a sphere. The center of the sphere lies on the plumb line at a distance from the support point and the radius of the sphere has the length ${\displaystyle {\tfrac {\Theta _{3}c_{0}}{(\Theta _{1}-\Theta _{3})L_{3}}}}$

${\displaystyle r={\sqrt {{\frac {2E}{\Theta _{1}}}+{\frac {2\Theta _{3}c_{0}L_{z}}{\Theta _{1}(\Theta _{1}-\Theta _{3})L_{3}}}+{\frac {\Theta _{3}^{2}c_{0}^{2}}{(\Theta _{1}-\Theta _{3})^{2}L_{3}^{2}}}+\left({\frac {1}{\Theta _{3}}}-{\frac {1}{\Theta _{1}}}\right){\frac {L_{3}^{2}}{\Theta _{3}}}}}}$

In it is

• Θ _{1 is} the equatorial moment of inertia,
• Θ ₃ the axial mass moment of inertia,
• L _{z is} the angular momentum around the plumb line,
• L _{3 is} the axial angular momentum around the figure axis ,
• c ₀  = mgs the support point moment, formed from the weight mg and the distance s of the center of mass from the support point on the figure axis, and
• E is the total mechanical energy of the top.

The distance from the support point to the center of the sphere and its radius increase beyond all limits when Θ ₁  = Θ ₃ , i.e. with the spherical  top , or when L ₃ = 0 and the top becomes a pendulum. What has been said about the Poinsot ellipsoid and the polhodia remains valid in these special cases.

Heavy spherical top

With the spherical top , the angular velocity and angular momentum are proportional to each other, which is why the angular velocity also moves in a plane that is perpendicular to the weight force.

Pendulum

In the case of the Lagrange gyro without axial angular velocity ω ₃ , the angular velocity and angular momentum are also proportional to one another, which is why what was said about the spherical gyroscope also applies here. A Lagrange top without an axial angular velocity ω ₃ therefore performs pendulum movements in which the end point of the angular velocity is in a horizontal plane.

Individual evidence

1. ^ Louis Poinsot: Théorie nouvelle de la rotation des corps. Bachelier, Paris 1834/1851, Grammel (1920), p. 24, Grammel (1950), p. 122 ff., Magnus (1971), p. 54, Leimanis (1965), p. 18, see literature.
2. ↑ Grammel (1920), p. 25
3. ↑ Grammel (1920), p. 24.
4. ↑ ^{a b} Grammel (1920), p. 36.
5. ↑ Grammel (1920), p. 35.
6. ↑ Léo Van Damme, Pavao Mardesic, Dominique Sugny: The tennis racket effect in a three-dimensional rigid body. June 28, 2016. Retrieved September 25, 2016 . 
7. ↑ Grammel (1920), p. 39.
8. ↑ Grammel (1920), p. 41.
9. ↑ Klein and Sommerfeld (2010), p. 217.
10. ↑ Klein and Sommerfeld (2010), p. 236.
11. ↑ Klein and Sommerfeld (2010), p. 201.

literature

• Louis Poinsot : New Theory of Rotation of Bodies . Bachelier, Paris 1834 (French, archive.org - original title: Théorie nouvelle de la rotation des corps .).
• R. Grammel : The top . Its theory and its applications. Vieweg Verlag, Braunschweig 1920, DNB  573533210 , p. 24, 32 ( archive.org - "swing" means angular momentum, " rotational shock " about the torque and "rotational force" rotational energy, see p VII). or R. Grammel : The top . Theory of the gyro. 2. revised Edition volume
  
   1 . Springer, Berlin, Göttingen, Heidelberg 1950, DNB  451641299 , p. 122 ff .
• K. Magnus : Kreisel: Theory and Applications . Springer, 1971, ISBN 978-3-642-52163-8 , pp. 53 ff . ( limited preview in Google Book Search [accessed January 7, 2020]).
• Eugene Leimanis: The General Problem of the Motion of Coupled Rigid Bodies about a Fixed Point . Springer Verlag, Berlin, Heidelberg 1965, ISBN 978-3-642-88414-6 , p. 18th ff ., doi : 10.1007 / 978-3-642-88412-2 (English, limited preview in Google Book Search [accessed January 7, 2020]).

Web links

• Svetoslav Zabunov: Stereo 3D Rigid Body Simulation. Zabunov Laboratories, accessed on October 11, 2016 (English, view set to "Poinsot construction (complete)").